电影院座位中的一个组合问题
A Combinatorial Problem in Cinema Seating
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中文总结 AI 辅助
本文研究电影院座位分配的组合问题，提出算法枚举满足条件的就座方案，并给出两种特殊情形下计数ψ(n)和ω(n)的递推公式及显式结果。
中文摘要 AI 辅助
我们研究一个关于电影院观众的问题：“一家电影院有编号为1到n的n个座位，有n个人持有编号为1到n的票。人们按顺序进入电影院。如果有人持有票号i，他们可以选择编号为i的倍数的座位。如果允许的座位已被先前的观众占据，他们应离开电影院。在这些条件下，他们有多少种就座方式？”我们给出一种算法来生成满足这些条件的情况列表。我们还专注于计算两种特殊情况下的情形数量：当恰好一个座位空着时，其总数记为ω(n)；以及当所有观众1,…,n-1都已就座时，其总数记为ψ(n)。给出递推公式ψ(n)=1+∑_{d|n, d≠n}ψ(d)，初始值ψ(1)=1，我们为ψ(p^α q^β)提供了显式公式，其中p和q是不同的素数。此外，我们证明ω(n)=-n+∑_{i=1}^nψ(i)。
英文摘要
We address a problem concerning cinema audiences: ``A cinema has $n$ seats numbered from $1$ to $n$, and there are $n$ people with tickets numbered from $1$ to $n$. People enter the cinema in order. If someone has the ticket number $i$, they can choose seats whose numbers are multiples of $i$. They should exit the cinema if the permitted seats are occupied by previous audience members. In how many ways can they be seated under these conditions?" We give an algorithm to create the list of situations that meet these conditions. We also focus on finding the number of situations in two special cases: when exactly one seat is unoccupied whose total number is denoted by $ω(n)$, and when all audiences $1, \ldots, n-1$ are seated, whose total number is denoted by $ψ(n)$. Giving the recursive formula $ψ(n)=1+\sum_{d|n, d\neq n}ψ(d)$ with the initial value $ψ(1)=1$, we provide an explicit formula for $ψ(p^αq^β)$, where $p$ and $q$ are distinct prime numbers. Furthermore, we show that $ω(n)=-n+\sum_{i=1}^nψ(i)$.
发表机构
- University of Torbat-e Jam（托尔巴特贾姆大学）
- Ferdowsi University of Mashhad（马什哈德费尔多西大学）
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